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  3. reaction is: pb(no3)2(aq) + 2ki(aq) ---> pbi2(s

Reaction is: Pb(NO3)2(aq) + 2KI(aq) ---> PbI2(s) + 2KNO3(aq) 2.0 ml of 0.250M Pb(NO3)2 and 7.0...

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Chemistry

Reaction is: Pb(NO3)2(aq) + 2KI(aq) ---> PbI2(s) + 2KNO3(aq)2.0 ml of 0.250M Pb(NO3)2 and 7.0 ml of 0.250M KI

A) In your experiment, one of thereagents was limiting and the other was in excess. Therefore afterthe lead(II)iodide precipitated from the solution, there werespectator ions left in the solution, and also the unreacted excessreagent. Assuming complete precipitation of lead (II) iodide,calculate the concentrations of each type of ion (Pb2+, K+, NO3-,I-) present in the solution after the PbI2 precipitatedout.

Answer & Explanation Solved by verified expert
4.2 Ratings (713 Votes)
Moles of PbNO32 20 mL x 0250 M 050 mmolMoles of KI 70 mL x 0250 M 175 mmolPbNO32aq 2KIaq PbI2s 2KNO3aqFor complete reaction the required mol ratio ofPbNO32 KI 1 2Here the available mol ratio 050 175 1 35Thus per    See Answer
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